Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 11
A Projectile thrown with the speed 100m/s. Making  an angle 60°. With the horizontal. Find the time after which the inclination is 45°.
7 months ago

Answers : (2)

Vikas TU
11392 Points
Dear stuent 
Initial and final velocities along the horizontal direction and vertical direction are :
ux = ucos60 
uy = usin60 
vx = vcos45 
vy = vsin45 
Using v = u + at 
ay = -9.8m/s^2 
ax = 0 m/s^2 
Vx = ux + axt 
v = u/√2 ........ (1)
Along vertical direction 
Vy = uy + ayt 
v/√2 = u√3/2 - 9.8t 
Put this in equation 1 
and find the vaue of t 
t = 3.73 sec 
h = uyt + 0.5ayt^2 
h = 18.43 m
Hope this helps
Good Luck 
7 months ago
24711 Points

let after time t its inclination with horizontal is 45

   at this time Vy /Vx =tan45  or Vy =Vx 

    Vx remains constant =100cos60 =Vy 

    now using equation of motion for verticle direction 


   Uy=100sin60 ,Uy=100cos60 

 putting in eq 1, 

    you can find t

7 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 18 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details